Answer :
(1) The mass of silver nitrate is, 555 g
(2) The solubility of the gas will be, 0.433 g/L
<u>Solution for Part 1 :</u>
From the given data we conclude that
In 100 gram of water, the amount of silver nitrate = 222 g
In 250 gram of water, the amount of silver nitrate = ![\frac{222}{100}\times 250=555g](https://tex.z-dn.net/?f=%5Cfrac%7B222%7D%7B100%7D%5Ctimes%20250%3D555g)
Therefore, the mass of silver nitrate is, 555 g
<u>Solution for Part 2 :</u>
Formula used :
(at constant temperature)
where,
= initial solubility of methane gas = 0.026 g/L
= final solubility of methane gas
= initial pressure of methane gas = 1 atm
= final pressure of methane gas = 0.06 atm
Now put all the given values in the above formula, we get the solubility of methane gas.
![(0.026g/L)\times (1atm)=S_2\times (0.06atm)](https://tex.z-dn.net/?f=%280.026g%2FL%29%5Ctimes%20%281atm%29%3DS_2%5Ctimes%20%280.06atm%29)
![S_2=0.433g/L](https://tex.z-dn.net/?f=S_2%3D0.433g%2FL)
Therefore, the solubility of the gas will be, 0.433 g/L